What is the integral of sqrt(1+((y-1)^2)/(4y)) ?

The integral of sqrt(1+((y-1)^2)/(4y)) is 1/3 sqrt(y)(y+3)+C

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integral of sqrt(1+((y-1)^2)/(4y))

\int 1 + \frac{( y - 1 ) ^{2}}{4 y} d y

\frac{1}{3} y (y + 3) + C

Solution steps

\int 1 + \frac{( y - 1 ) ^{2}}{4 y} d y

Simplify

1 + \frac{( y - 1 ) ^{2}}{4 y} : \frac{( y + 1 ) ^{2}}{2 y}

= \int \frac{( y + 1 ) ^{2}}{2 y} d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{( y + 1 ) ^{2}}{y} d y

(y + 1)^{2} = ((y + 1)),

assuming

(y + 1) \geq 0

= \frac{1}{2} \cdot \int \frac{y + 1}{y} d y

Expand

\frac{y + 1}{y} : y + \frac{1}{y}

= \frac{1}{2} \cdot \int y + \frac{1}{y} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int y d y + \int \frac{1}{y} d y)

\int y d y = \frac{2}{3} y^{\frac{3}{2}}

\int \frac{1}{y} d y = 2 y

= \frac{1}{2} (\frac{2}{3} y^{\frac{3}{2}} + 2 y)

Simplify

\frac{1}{2} (\frac{2}{3} y^{\frac{3}{2}} + 2 y) : \frac{1}{3} y (y + 3)

= \frac{1}{3} y (y + 3)

Add a constant to the solution

= \frac{1}{3} y (y + 3) + C

What is the integral of sqrt(1+((y-1)^2)/(4y)) ?
The integral of sqrt(1+((y-1)^2)/(4y)) is 1/3 sqrt(y)(y+3)+C